1/n now is a problem

Is it possible, that if a 1 < a 2 < a 3 < a 4 < < a 2017 a_1<a_2<a_3<a_4<\dots<a_{2017} are positive integers, then 1 a 1 1 a 2 = 1 a 2 1 a 3 = 1 a 3 1 a 4 = = 1 a 2016 1 a 2017 ? \dfrac{1}{a_1}-\dfrac{1}{a_2}=\dfrac{1}{a_2}-\dfrac{1}{a_3}=\dfrac{1}{a_3}-\dfrac{1}{a_4}=\dots =\dfrac{1}{a_{2016}}-\dfrac{1}{a_{2017}}?

No, it is not possible Yes, it is possible

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1 solution

Sándor Daróczi
Sep 9, 2017

Define the sequence a n a_n by the following:

a n = k N n , 1 n < N a_n=\frac{k}{N-n}, \ 1 \leq n<N

where k k and N N are fixed positive integers.

We'll prove that for some appropiate k k and N N this sequence fulfills the problem's conditions.

First, we seek to show that the difference 1 a i 1 a i + 1 \frac{1}{a_i}-\frac{1}{a_{i+1}} does not depend on the value of i i . Indeed,

1 a i 1 a i + 1 = N i k N i 1 k = 1 k \frac{1}{a_i}-\frac{1}{a_{i+1}} = \frac{N-i}{k}-\frac{N-i-1}{k} = \frac{1}{k} .

Letting k = N ! k=N! it follows that N n k {N-n} | k for every 1 n < N 1 \leq n<N , hence the value of a n = k N n a_n=\frac{k}{N-n} is a positive integer. Since the condition a i < a i + 1 a_i<a_{i+1} is the simple consequence of the fact that k N i < k N i 1 \frac{k}{N-i} < \frac{k}{N-i-1} , we have shown that this sequence is a correct solution for any N N . Putting in N = 2018 N=2018 we are done.

I got the same solution!

Áron Bán-Szabó - 3 years, 9 months ago

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